Prove that $ { \lim_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = e^{-1} } $. I want to prove that $$  { \lim_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = 
 e^{-1} } $$. I came up with a proof, but want to make sure that it is correct. Here is my proof:
$$  { \lim_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = \lim_{n\to\infty} \left(\frac{n-1}{n} \right)^n = \lim_{n\to\infty} \left(\frac{n}{n-1} \right)^{-n} = \\
\lim_{n\to\infty} \left((1+\frac{1}{n-1} )^n\right)^{-1} = \lim_{n-1\to\infty} \left((1+\frac{1}{n-1} )^{n-1}(1+\frac{1}{n-1})\right)^{-1} = \\
\lim_{n-1\to\infty} \left(e(1+\frac{1}{n-1})\right)^{-1} = e^{-1}}  $$
Is it valid proof? I know there are other proofs, but I want to know about this one. In particular can we substitude $e$ instead of $(1+\frac{1}{n-1} )^{n-1}$ and $1$ instead of $(1+\frac{1}{n-1})$ in the limit?
Thank you for your answers.
 A: Your proof idea is correct, but needs to be written a little better. Let us see how we can do this.
First of all, you start off with manipulating the limit $\lim_{n \to \infty} \left(1 - \frac 1n\right)^n$, without knowing that it exists. Sure, it does exist in the end, but to manipulate it you need to know that it exists, and is a real number, right? Unless you justify this prior to the manipulations, you will be in the wrong.
That is, your proof steps are valid only if you show that $\lim_{n \to \infty} \left(1 - \frac 1n\right)^n$ exists before doing all this. You have not done this, and hence I would not call your proof complete.
However, by simply reversing your steps, you can ensure that you don't need to provide proof of the limit existence, but just get its direct value, and therefore its existence. We'll see why.
To see this, start off  with $e = \lim_{n \to \infty} \left(1+\frac 1n\right)^n$, which you must have done before the exercise.
We shift $ n \to n-1$ to get $\lim_{n \to \infty} \left(\frac n{n-1}\right)^{n-1} =e $.  
Taking the reciprocal of the limit gives us $\lim_{n \to \infty} \left(\frac{n-1}{n}\right)^{n-1} = e^{-1}$(We are using $e \neq 0$ here, but you should know this).
The limit $\lim_{n \to \infty}\frac {n-1}n = 1$ is easy to see. By the product of limits result, taking the product with the previous line gives $\lim_{n \to \infty} \left(\frac{n-1}{n}\right)^n = e^{-1}$.
By separating $\frac{n-1}n = 1 - \frac 1n$ above, you can see that the desired limit exists, and its value is obtained without having to manipulate  or use it at all. This proof has essentially rearranged your steps in such a manner that you are not using manipulating the limit you want to calculate, prior to finding its value, without showing its existence. Your proof would have worked, if you had showed that $\lim_{n \to \infty} \left(\frac{n-1}{n}\right)^n$ exists before doing all the manipulation that you did.
So, keep in mind that manipulating a limit requires you to prove its existence. If you cannot do this, then a "reversal of steps" like I have done can be helpful.
EDIT : I am leaving the above proof as a "model proof" for this problem. However, as a comment below points out, all the manipulations that you have performed (taking reciprocals, separating terms etc.) have been performed inside the limit, which is completely permissible. However, just because there is a limit sitting behind each of these expressions, and in particular the first expression, whose limit you are to find, it may still appear that you are assuming that the limit exists.
For example, consider the following clearly incorrect sequence of statements :
$$
\lim_{x \to 0} \frac 1x = \lim_{x \to 0} x^{-1} = \lim_{x \to 0} x \times x^{-2}
$$
Since the first limit does not exist, neither does any of the others. So the equalities are false.
However, suppose we remove the "$\lim_{x \to 0}$" part :
$$
\frac 1x = x^{-1} = x \times x^{-2}
$$
then these are perfectly true statements (for $x \neq 0$). Something like this , if done to your proof, massively improves it.
That is to say, in your proof, just remove "$\lim$" from all the expressions except the last. So now your proof looks like :
$$
\left(1 - \frac 1n\right)^n = ... = \left( \left( 1 + \frac 1{n-1}\right)^{n-1}\left(1 + \frac 1{n+1}\right)\right)^{-1}
$$
and then say : the limit of $\left(1+ \frac 1{n+1}\right)^{n-1}$ exists and equals $e$. Similarly, the limit of $\left(1 + \frac 1{n+1}\right)$ exists and equals $1$. Now, the limit of the right hand side exists and equals $e^{-1}$ from the product and reciprocal rule for limits. Thus the limit of the left hand side also exists and equals $e^{-1}$. 
So I think just the fact that you place "$\lim$" while doing the algebraic manipulations, is causing trouble. Remove it, and the trouble is as good as gone.
A: Your proof is correct, however as stated in other answers it is best not to talk about the limit of an expression if you have not proved said limit exists. Otherwise it is a circular argument: "Assuming the limit exists, I can prove that the limit exists [and is equal to...]".
So the writing should rather be 
$${\left( 1-\frac{1}{n} \right)^n = \left(\frac{n-1}{n} \right)^n =\left(\frac{n}{n-1} \right)^{-n} = \\
 \left((1+\frac{1}{n-1} )^n\right)^{-1} =  \left((1+\frac{1}{n-1} )^{n-1}(1+\frac{1}{n-1})\right)^{-1}}$$
Then, you use the known fact that $\displaystyle\lim_{n\to \infty} \left(1 + \frac{1}{n}\right)^n = e$, that $\displaystyle\lim_{n\to \infty} 1 + \frac{1}{n-1} = 1$, so by product and multiplication of limits, the limit of $\left( 1-\frac{1}{n} \right)^n$ exists and is equal to $\frac 1e$.
